What is the general solution for 9cos^2(x)-18cos(x)+5=0 ?

The general solution for 9cos^2(x)-18cos(x)+5=0 is x=1.23095…+2pin,x=2pi-1.23095…+2pin

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9cos^2(x)-18cos(x)+5=0

Solution

9 cos^{2} (x) - 18 cos (x) + 5 = 0

Solution

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn

Degrees

x = 70.52877 \dots^{\circ} + 36 0^{\circ} n, x = 289.47122 \dots^{\circ} + 36 0^{\circ} n

Solution steps

9 cos^{2} (x) - 18 cos (x) + 5 = 0

Solve by substitution

9 cos^{2} (x) - 18 cos (x) + 5 = 0

Let:

cos (x) = u

9 u^{2} - 18 u + 5 = 0

9 u^{2} - 18 u + 5 = 0 : u = \frac{5}{3}, u = \frac{1}{3}

9 u^{2} - 18 u + 5 = 0

Solve with the quadratic formula

9 u^{2} - 18 u + 5 = 0

Quadratic Equation Formula:

For

a = 9, b = - 18, c = 5

u_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 9 \cdot 5}{2 \cdot 9}

u_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 9 \cdot 5}{2 \cdot 9}

(- 18)^{2} - 4 \cdot 9 \cdot 5 = 12

(- 18)^{2} - 4 \cdot 9 \cdot 5

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 18)^{2} = 1 8^{2}

= 1 8^{2} - 4 \cdot 9 \cdot 5

Multiply the numbers:

4 \cdot 9 \cdot 5 = 180

= 1 8^{2} - 180

1 8^{2} = 324

= 324 - 180

Subtract the numbers:

324 - 180 = 144

= 144

Factor the number:

144 = 1 2^{2}

= 1 2^{2}

Apply radical rule:

1 2^{2} = 12

= 12

u_{1, 2} = \frac{- ( - 18 ) \pm 12}{2 \cdot 9}

Separate the solutions

u_{1} = \frac{- ( - 18 ) + 12}{2 \cdot 9}, u_{2} = \frac{- ( - 18 ) - 12}{2 \cdot 9}

u = \frac{- ( - 18 ) + 12}{2 \cdot 9} : \frac{5}{3}

\frac{- ( - 18 ) + 12}{2 \cdot 9}

Apply rule

- (- a) = a

= \frac{18 + 12}{2 \cdot 9}

Add the numbers:

18 + 12 = 30

= \frac{30}{2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{30}{18}

Cancel the common factor:

6

= \frac{5}{3}

u = \frac{- ( - 18 ) - 12}{2 \cdot 9} : \frac{1}{3}

\frac{- ( - 18 ) - 12}{2 \cdot 9}

Apply rule

- (- a) = a

= \frac{18 - 12}{2 \cdot 9}

Subtract the numbers:

18 - 12 = 6

= \frac{6}{2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{6}{18}

Cancel the common factor:

6

= \frac{1}{3}

The solutions to the quadratic equation are:

u = \frac{5}{3}, u = \frac{1}{3}

Substitute back

u = cos (x)

cos (x) = \frac{5}{3}, cos (x) = \frac{1}{3}

cos (x) = \frac{5}{3}, cos (x) = \frac{1}{3}

cos (x) = \frac{5}{3} :

No Solution

cos (x) = \frac{5}{3}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{1}{3} : x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

cos (x) = \frac{1}{3}

Apply trig inverse properties

cos (x) = \frac{1}{3}

General solutions for

cos (x) = \frac{1}{3}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

Combine all the solutions

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

Show solutions in decimal form

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 9cos^2(x)-18cos(x)+5=0 ?
The general solution for 9cos^2(x)-18cos(x)+5=0 is x=1.23095…+2pin,x=2pi-1.23095…+2pin